TITLE: finite index subgroup of a fuchsian group
QUESTION [5 upvotes]: Given G, a fuchsian group and a finite sub set A of G. Does there exist a finite index subgroup H in G such that inter section of A with H is empty?

REPLY [12 votes]: On Yves' request, here is a proof of residual finiteness of non-finitely generated Fuchsian groups. (I do not know a reference for this: In general, people do not like working with infinitely generated groups, so, it is possible that nobody bothered writing a proof, even though many people could. The result could be in Beardon's or Maskit's book though, I did not check.) A side remark: Residual finiteness of Fuchsian groups was a partial motivation for this question. 
A Fuchsian group is a discrete group $\Phi$ of isometries of hyperbolic plane $H^2$. Then $\Phi$ contains an index 2 subgroup $\Gamma$ which preserves orientation on the hyperbolic plane and, hence, $\Gamma < PSL(2, R)$. Since residual finiteness is a commensurablity invariant, it suffices to consider the case of discrete subgroups of $PSL(2,R)$. Let $O=H^2/\Gamma$  be the quotient orbifold. Its singular set is a discrete subset $\Sigma=\{x_i: i\in I\}$ of the underlying space $X$ of $O$. I will assume that $\Gamma$ is not finitely generated, thus, $O$ is noncompact.  
Let $T\subset X$ be an infinite locally finite, 1-ended properly embedded tree (with geodesic edges) which contains $\Sigma$. Let $U$ be a small neighborhood (with smooth boundary) of $T$ in $X$ which admits a  retraction to $T$. Every boundary component of $U$ is contractible. Let $O_U$ be the suborbifold of $O$ supported by $U$. Thus, by Seifert-van Kampen theorem for orbifolds applied to this situation, the fundamental group $\Gamma$ of $O$ splits as a free product 
$$
\pi_1(O_U) * \prod^*_i \pi_1(S_i)
$$ 
where $S_i$'s are components of $X\setminus U$, and $\prod^*$ means free product. Now, since each surface $S_i$ is noncompact, by a theorem of Whitehead, it is homotopy-equivalent to a graph (see the discussion here). Hence, $\Gamma \cong  \pi_1(O_U) * F$, where $F$ is a free group. Lastly (again by Seifert-van Kampen theorem applied to $O_U$), the group $\pi_1(O_U)$ is a free product of finite cyclic groups $Z_{n_i}$, where $n_i$ is the order of the singular point $x_i\in \Sigma$. Thus, $\Gamma$ is the free product of countably many cyclic groups (some of which might be finite  and some infinite). 
The converse is also true: If $\Gamma$ is a countable product of cyclic groups $C_i$, then $\Gamma$ is isomorphic to a discrete subgroup of $PSL(2, R)$. Namley,  make each $C_i$ act 
isometrically on $H^2$ with the fundamental domain $D_i$, so that the sets $cl(H^2 -D_i)$ are pairwise disjoint. Now, take the subgroup $\Phi$ of $PSL(2, R)$ generated by the subgroups $C_i$. By Klein combination theorem (also known as the "ping-pong argument") $\Phi$ is discrete (its fundamental domain is the intersection of $D_i$'s) and isomorphic to $\Gamma$. 
Now, suppose that $\Gamma$ is a group which is a countable free product of residually finite groups $\Gamma_i, i\in {\mathbb N}$. Let $S\subset \Gamma$ be a finite subset not containing the identity. Then 
$$
S\subset \prod^*_{j\in J} \Gamma_j$$
where $J\subset I$ is a finite subset. Hence, $S$ projects bijectively to a subset $P$ in the quotient
$$
\Lambda=\Gamma/\langle \langle \bigcup_{i\notin J} \Gamma_i\rangle \rangle. 
$$
Now, $\Lambda$ is residually finite (as a finite free product of residually finite groups). Hence, $\Lambda$ contains a finite-index subgroup $\Lambda'$ disjoint from $P$. Taking preimage of $\Lambda'$ under the epimomorphism $\Gamma \to \Lambda$, we obtain a finite index subgroup in $\Gamma$ disjoint from $S$. This argument, of course, applies to free products of cyclic groups, thereby proving residual finiteness of non-finitely generated Fuchsian groups.